The probability that a seed of a certain variety germinates is 0.8. If 4 such seeds are sown, the probability that exactly 3 of them germinate is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.4096
Let X be the number of seeds that germinate. The seeds behave independently, so X follows a binomial distribution with n = 4, p = 0.8 and q = 0.2:
\(P(X = r) = {}^{n}C_{r}\,p^{r}\,q^{\,n - r}\)
For r = 3:
\(P(X = 3) = {}^{4}C_{3}\,(0.8)^{3}\,(0.2)^{1}\)
\(= 4 \times 0.512 \times 0.2\)
\(= 0.4096\)
Why the other options are incorrect
Option (1) 0.512 — Incorrect. This is only \((0.8)^{3}\); the factor 0.2 for the seed that fails and the 4 ways of choosing it are missing.
Option (2) 0.8192 — Incorrect. This is \(P(X = 3) + P(X = 4) = 0.4096 + 0.4096\), the probability that at least 3 seeds germinate.
Option (3) 0.1536 — Incorrect. This is \(P(X = 2) = 6 \times 0.64 \times 0.04\), exactly 2 seeds germinating.